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Ta có:

1,(454) =  0,(001) . 454 + 1 = \(\frac{1}{999}.454+1\)= \(\frac{454}{999}+1=\frac{1453}{999}\)

1,4(545) = 0,0(001) . 545 + 1,4 = \(\frac{1}{9990}.545+1,4=\frac{109}{1998}+1,4=\frac{14531}{9990}\)

=> 1,(454) < 1,4(545)

0,(428571) = 428571.0,(000001)

a,A=0,(000001).48571=1/999999.428571=3/7=B

b,C=0,(01).17+0,(01).82=0,(01).(17+82)=99/99=1=D

c,D=2³³²<2³³³=8¹¹¹<9¹¹¹=3²²²<3²²³

d,4F=1+1/4...+1/4⁹⁹⁹

=>4F-F=1-1/4¹⁰⁰⁰

=>3F=1-1/4¹⁰⁰⁰<1

=>F<1/3=>F<G

Bài 1:

a) \(\dfrac{-17}{36}\) và \(\dfrac{23}{-48}\) 

\(\dfrac{-17}{36}=\dfrac{-17.4}{36.4}=\dfrac{-68}{144}\) 

\(\dfrac{23}{-48}=\dfrac{-23}{48}=\dfrac{-23.3}{144.3}=\dfrac{-69}{144}\) 

Vì \(\dfrac{-68}{144}>\dfrac{-69}{144}\) nên \(\dfrac{-17}{36}>\dfrac{23}{-48}\) 

b) \(\dfrac{-1}{3}\) và \(\dfrac{2}{5}\) 

Vì \(\dfrac{-1}{3}\) là số âm mà \(\dfrac{2}{5}\) là số dương nên \(\dfrac{-1}{3}< \dfrac{2}{5}\) 

c) \(\dfrac{2}{7}\) và \(\dfrac{5}{4}\) 

Vì \(\dfrac{2}{7}< 1\) mà \(\dfrac{5}{4}>1\) nên \(\dfrac{2}{7}< \dfrac{5}{4}\) 

d) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\) 

\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\) 

\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\) 

Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

Bài 2:

\(\dfrac{5}{2}-\left(1\dfrac{3}{7}-0,4\right)=\dfrac{5}{2}-\dfrac{10}{7}-\dfrac{2}{5}=\dfrac{47}{70}\) 

a) Ta có: 1,(81) = 1,8181…

Vì 1,8181… > 1,812 nên -1,8181… < -1,812 hay -1,(81) < -1,812

b) Ta có: \(2\frac{1}{7}\) = 2,142857….

Vì 2,142857….> 2,142 nên \(2\frac{1}{7}\) > 2,142

c) Vì 48,075… < 48,275… nên - 48,075…. > – 48,275…

d) Vì 5 < 8 nên \(\sqrt 5 \) < \(\sqrt 8 \)

a: -1,(81)>-1,812

b: 2+1/7>2,142

c: -48,075...>-48,275...

d: \(\sqrt{5}< \sqrt{8}\)

a)

\(\begin{array}{l}{( - 3)^2}.{( - 3)^4} = 9.81 = 729\\ {( - 3)^6} = ( - 3).( - 3).( - 3).( - 3).( - 3).( - 3)\\ = 9.9.9 = 729\end{array}\)

Vậy \({( - 3)^2}.{( - 3)^4}\) = \({( - 3)^{6}}\)

b)

\(\begin{array}{l}0,6{}^3:0,{6^2} = 0,216:0,36 = 0,6\end{array}\)

Vậy \(0,6{}^3:0,{6^2}\) = \(0,{6}\)